core mathematics c4 (6666) level... · 6666/01 core maths c4 june 2006 advanced subsidiary/...

GCE Edexcel GCE

Core Mat hemat ics C4 (6666)

June 2006

Mark Scheme

(Final)

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6666/01 Core Maths C4 June 2006 Advanced Subsidiary/ Advanced Level in GCE Mat hemat ics

2

June 2006

6666 Pure Mathematics C4

Mark Scheme

Question Number

Scheme Marks

1.

⎧ ⎫= − + − =⎨ ⎬

⎩ ⎭

dy dy dy6x 4y 2 3 0

dx dx dx

Differentiates implicitly to include either

dyky

dx± or

dy3

dx± . (Ignore

⎛ =⎜ ⎟⎝

dy

dx

⎞⎠

.)

Correct equation.

M1 A1

dy 6x 2

dx 4y 3

⎧ ⎫+=⎨ ⎬

+⎩ ⎭ not necessarily required.

At (0, 1), +

= =+

dy 0 2 2

dx 4 3 7

Substituting x = 0 & y = 1 into an equation

involving dy

dx;

to give 27

or 27−−

dM1; A1 cso

Hence m(N) = −7

2 or

27

1− Uses m(T) to ‘correctly’ find m(N). Can be

ft from “their tangent gradient”.A1 oe.

Either N: − = − −72

y 1 (x 0)

or N: 72

y x= − + 1

y 1 m(x 0)− = − with

‘their tangent or normal gradient’;or uses y mx 1= + with ‘their tangent or

normal gradient’ ;

M1;

N: 7x + 2y – 2 = 0 Correct equation in the form

+ + ='ax by c 0 ' ,

where a, b and c are integers.

A1 oe cso

[7]

7 marks

Beware: dy 2

dx 7= does not necessarily imply the award of all the first four marks in this question.

So please ensure that you check candidates’ initial differentiation before awarding the first A1 mark. Beware: The final accuracy mark is for completely correct solutions. If a candidate flukes the final line then they must be awarded A0. Beware: A candidate finding an m(T) = 0 can obtain A1ft for m(N) = ∞ , but obtains M0 if they write

. If they write, however, N: x = 0, then can score M1. y 1 (x 0)− = ∞ − Beware: A candidate finding an m(T) = can obtain A1ft for m(N) = 0, and also obtains M1 if they write

or y = 1. ∞

y 1 0(x 0)− = − Beware: The final cso refers to the whole question.


3

Question Number

Scheme Marks

Aliter

1.

d x

d y

dx dx6x 4y 2 3 0

dy dy

⎧ ⎫⎪ ⎪= − + −⎨ ⎬⎪ ⎪⎩ ⎭

=

Differentiates implicitly to include either

dxkx

dy± or

dx2

dy± . (Ignore

dx

dy

⎛=⎜ ⎟

⎝

⎞

⎠.)

Correct equation.

M1 A1

Way 2

dx 4y 3

dy 6x 2

⎧ ⎫+=⎨ ⎬

+⎩ ⎭ not necessarily required.

At (0, 1), dx 4 3 7

dy 0 2 2

+= =

+

Substituting x = 0 & y = 1 into an

equation involving dxdy

;

to give 72

dM1; A1 cso

Hence m(N) = −7

2 or

27

1−

Uses m(T) or dxdy

to ‘correctly’ find m(N).

Can be ft using “ dxdy

1.− ”.A1 oe.

Either N: − = − −72

y 1 (x 0)

or N: 72

y x= − + 1

y 1 m(x 0)− = − with

‘their tangent, dxdy

or normal gradient’;

or uses y mx 1= + with ‘their tangent, dxdy

or normal gradient’ ;

M1;

N: 7x + 2y – 2 = 0 Correct equation in the form

+ + ='ax by c 0 ' ,

where a, b and c are integers.

A1 oe cso

7 marks


4

Question Number

Scheme Marks

Aliter

1. + − − − =2 22y 3y 3x 2x 5 0

Way 3

( )+ − = + +223 9 3x4 16 2

y x 52

( )= + + −23x 49 32 16

y x4

( ) ( )−

= + + +1

2 23x 492 16

dy 1x 3x

dx 21

Differentiates using the chain rule;

Correct expression for dy

dx.

M1; A1 oe

At (0, 1), −

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

12dy 1 49 1 4 2

dx 2 16 2 7 7=

Substituting x = 0 into an equation involving dy

dx;

to give 27

or 27−−

dM1 A1 cso

Hence m(N) = −7

2

Uses m(T) to ‘correctly’ find m(N).Can be ft from “their tangent gradient”.

A1

Either N: − = − −72

y 1 (x 0)

or N: 27

y x= − + 1

y 1 m(x 0)− = − with

‘their tangent or normal gradient’;or uses y mx 1= + with ‘their tangent or

normal gradient’

M1

N: 7x + 2y – 2 = 0 Correct equation in the form ,+ + ='ax by c 0 '

where a, b and c are integers.A1 oe

[7]

7 marks


5

Question Number

Scheme Marks

2. (a) − ≡ − +3x 1 A(1 2x) B

Considers this identity and either substitutes

12

x = , equates

coefficients or solves simultaneous

equations

complete M1

Let 12

x ;= − = ⇒ =3 12 2

1 B B

Equate x terms; 32

3 2A A= − ⇒ = − 3 12

A ; B= − =2 A1;A1

(No working seen, but A and B correctly stated ⇒ award all

three marks. If one of A or B correctly stated give two out of the three marks available for this part.)

[3]

(b) − −= − − + −1 23 1

2 2f(x) (1 2x) (1 2x)

Moving powers to top on any one of the two

expressionsM1

23

2

( 1)( 2) ( 1)( 2)( 3)1 ( 1)( 2x); ( 2x) ( 2x) ...

2! 3!

⎧ ⎫− − − − −= − + − − + − + − +⎨ ⎬

⎭⎩3

Either 1 or 1 from either first or

second expansions respectively

2x± 4x±

dM1;

21

2

( 2)( 3) ( 2)( 3)( 4)1 ( 2)( 2x); ( 2x) ( 2x) ...

2! 3!

⎧ ⎫− − − − −+ + − − + − + − +⎨ ⎬

⎭⎩3

Ignoring 32

− and 12

,

any one correct

}{.......... expansion.

Both }{.......... correct.

A1 A1

}{ }{= − + + + + + + + + +2 3 2 33 1

2 21 2x 4x 8x ... 1 4x 12x 32x ...

2 31 x ; 0x 4x= − − + + 21 x ; (0x ) 4x− − + 3

A1; A1

[6] 9 marks

Beware: In part (a) take care to spot that 32

A = − and 12

B = are the right way around.

Beware: In ePEN, make sure you aware the marks correctly in part (a). The first A1 is for 32

A = − and the

second A1 is for 12

B = .

Beware: If a candidate uses a method of long division please escalate this to you team leader.


6

Question Number

Scheme Marks

Aliter

2. (b) −= − − 2f(x) (3x 1)(1 2x) Moving power to top M1

Way 2

( )2

3

( 2)( 3)1 ( 2)( 2x) ; ( 2x)

2!3x 1

( 2)( 3)( 4)( 2x) ...

3!

− −⎛ ⎞+ − − + − +⎜ ⎟⎜ ⎟= − ×

− − −⎜ ⎟− +⎜ ⎟⎝ ⎠

1 4x± ; Ignoring ( , correct3x 1)−

( )........... expansion

dM1; A1

= − + + + +2 3(3x 1)(1 4x 12x 32x ...)

2 3 2 33x 12x 36x 1 4x 12x 32x ...= + + − − − − + Correct expansion A1

2 31 x ; 0x 4x= − − + + 21 x ; (0x ) 4x− − + 3

A1; A1

[6]

Aliter 2. (b) Maclaurin expansion Way 3

− −= − − + −1 23 12 2

f(x) (1 2x) (1 2x) Bringing both powers to top

M1

2 3f (x) 3(1 2x) 2(1 2x)− −′ = − − + −

Differentiates to give

;2 3a(1 2x) b(1 2x)− −− ± −2 33(1 2x) 2(1 2x)− −− − + −

M1; A1 oe

3 4f (x) 12(1 2x) 12(1 2x)− −′′ = − − + −

4 5f (x) 72(1 2x) 96(1 2x)− −′′′ = − − + − Correct f (x) and f (x)′′ ′′′ A1

f(0) 1 , f (0) 1 , f (0) 0 and f (0) 24′ ′′ ′′′∴ = − = − = =

2 3gives f(x) 1 x; 0x 4x ...= − − + + + 21 x ; (0x ) 4x− − + 3

A1; A1

[6]


7

Question Number

Scheme Marks

Aliter

2. (b) 1 21

2f(x) 3(2 4x) (1 2x)− −= − − + −

Moving powers to top on any one of the two

expressionsM1

Way 4

1 2 3

4 3

( 1)( 2)(2) ( 1)(2) ( 4x); (2) ( 4x)

2!3

( 1)( 2)( 3)(2) ( 4x) ...

3!

− − −

−

⎧ ⎫− −+ − − + −⎪ 2

⎪⎪ ⎪= − ⎨ − − −⎪ ⎪+ − +⎪ ⎪⎭⎩

⎬

Either 12

x± or 1

from either first or second expansions

respectively

4x±

dM1;

21

2

( 2)( 3) ( 2)( 3)( 4)1 ( 2)( 2x); ( 2x) ( 2x) ...

2! 3!

⎧ ⎫− − − − −+ + − − + − + − +⎨ ⎬

⎭⎩3

Ignoring and 3− 12

,

any one correct

}{.......... expansion.

Both }{.......... correct.

A1 A1

}{ }{2 3 2 31 1

2 23 x 2x 4x ... 1 4x 12x 32x ...= − + + + + + + + + +

2 31 x ; 0x 4x= − − + + 21 x ; (0x ) 4x− − + 3

A1; A1

[6]


8

Question Number

Scheme Marks

3. (a) Area Shaded = ( )π

∫2

x2

0

3sin dx

( ) π

⎡ ⎤−= ⎢ ⎥⎣ ⎦

2x2

12 0

3cos

Integrating ( )x2

3sin to give

( )x2

k cos with .≠k 1

Ignore limits.

M1

( ) π⎡ ⎤= −⎣ ⎦

2x2 0

6cos ( )− x2

6cos or ( )12

3 x2

cos− A1 oe.

[ ] [ ]= − − − − = + =6( 1) 6(1) 6 6 12 12 A1 cao

[3]

(Answer of 12 with no working scores M0A0A0.)

(b) Volume ( )( ) ( )2 2

2 2x x2 2

0 0

3sin dx 9 sin dx

π π

= π = π∫ ∫ Use of 2V y d= π∫ x .

Can be implied. Ignore limits.

M1

1 cos2x2 2

2NB : cos2x 1 2sin x gives sin x

−⎡ ⎤= ± ± =⎣ ⎦

( ) ( ) 1 cosx2 2x x2 2

NB : cos x 1 2sin gives sin−⎡ ⎤= ± ± =⎣ ⎦2

Consideration of the Half Angle

Formula for ( )2 x2

sin or the

Double Angle Formula

for 2sin x

M1∗

∴Volume

2

0

1 cos x9( ) dx

2

π−⎛ ⎞= π ⎜ ⎟

⎝ ⎠∫ Correct expression for Volume

Ignore limits and π .A1

( ) 2

0

9(1 cos x) dx

2

ππ

= −∫

( ) [ ]2

0

9x sin x

2

ππ= −

Integrating to give ax bsinx± ± ;

Correct integrationk k cos x kx k sin x− → −

depM1∗ ; A1

[ ]π= π − − −

9(2 0) (0 0)

2

π

= π9

(2 )2

= 9 π2 or 88.8264… Use of limits to give

either 9 π2 or awrt 88.8A1 cso

Solution must be completely correct. No flukes allowed.

[6]

9 marks


9

Question 3 Note: is not needed for the middle four marks of question 3(b). π Beware: Owing to the symmetry of the curve between x = 0 and x = 2π candidates can find:

• Area = 2 ( )x2

0

3sin dx

π

∫ in part (a).

• Volume = 2 ( )( )2x2

0

3sin dx

π

π ∫

Beware: If a candidate gives the correct answer to part (b) with no working please escalate this response up to your team leader.


10

Question Number

Scheme Marks

4. (a) , x sin t= ( )6y sin t π= +

dx

cos tdt

= , ( )6

dycos t

dtπ= +

Attempt to differentiate both x and y wrt t to give two terms in cos

Correct dxdt

and dy

dt

M1 A1

When t ,6

π=

( )( )

16 6 2

36 2

cosdy 1awrt 0.58

dx cos 3

π π

π

+= = = =

Divides in correct way and substitutes for t to give any of the

four underlined oe: Ignore the double negative if candidate has differentiated

sin cos→ −

A1

When t ,6

π=

31x , y

2 2= = The point ( )31

2 2, or ( )1

2, awrt 0.87 B1

T: ( )− = −3 1 12 23

y x

Finding an equation of a tangent with their point and their tangent

gradient or finds c and uses y (their gradient)x "c "= + .

Correct EXACT equation of tangent oe.

dM1 A1 oe

or ( )3 31 12 2 2 63

c c= + ⇒ = − =3 3

3

or T: 3 3

3 3y x⎡ ⎤= +⎣ ⎦

[6]

(b) ( )6 6y sin t sin t cos cos t sinπ π= + = +

6π Use of compound angle formula

for sine.M1

2 2 2Nb : sin t cos t 1 cos t 1 sin t+ ≡ ⇒ ≡ − 2

∴ x sin t= gives ( )2cos t 1 x= − Use of trig identity to find in

terms of x or in terms of x.

cos t2cos t

M1

3 1

2 2y sin t cos∴ = + t

gives ( )3 212 2

y x 1 x= + − AG

Substitutes for

6 6sin t , cos , cos t and sinπ π to

give y in terms of x.

A1 cso

[3]

9 marks


11

Question Number

Scheme Marks

Aliter

4. (a) , x sin t= ( )6 6y sin t sin t cos cos t sinπ π= + = +

6π (Do not give this for part (b))

Way 2

dxcos t

dt= ,

6 6

dycos t cos sin t sin

dtπ π= −

Attempt to differentiate x and y

wrt t to give dxdt

in terms of cos

and dy

dtin the

form acos t bsin t± ±

Correct dxdt

and dy

dt

M1 A1

When t ,6

π=

( )6 6 6 6

6

cos cos sin sindy

dx cos

π π π π

π

−=

3 1 14 4 2

3 3

2 2

1awrt 0.58

3

−= = = =

Divides in correct way and substitutes for t to give any of the

four underlined oe: A1

When t ,6

π=

31x , y

2 2= =

The point ( )312 2,

or ( )12, awrt 0.87

B1

T: ( )− = −3 1 12 23

y x




dM1 A1 oe

or ( )3 31 12 2 2 63

c c= + ⇒ = − =3 3

3

or T: 3 3

3 3y x⎡ ⎤= +⎣ ⎦

[6]


12

Question Number

Scheme Marks

Aliter

4. (a) ( )3 212 2

y x 1 x= + −

Way 3

( ) ( )122dy 3 1 1

1 x 2xdx 2 2 2

−⎛ ⎞ ⎛ ⎞= + − −⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

Attempt to differentiate two terms using the chain rule for the second

term.

Correct dy

dx

M1 A1

( ) ( )122dy 3 1 1 1

1 (0.5) 2(0.5)dx 2 2 2 3

−⎛ ⎞ ⎛ ⎞= + − − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

Correct substitution of 12

x =

into a correct dy

dx

A1

When t ,6

π=

31x , y

2 2= = The point ( )31

2 2, or ( )1

2, awrt 0.87 B1

T: ( )− = −3 1 12 23

y x




dM1 A1 oe

or ( )3 31 12 2 2 63

c c= + ⇒ = − =3 3

3

or T: 3 3

3 3y x⎡ ⎤= +⎣ ⎦

[6]

Aliter

4. (b) x sin t= gives ( )3 212 2

y sin t 1 sin= + − t Substitutes into the

equation give in y.

x sin t=M1

Way 2

2 2 2Nb : sin t cos t 1 cos t 1 sin t+ ≡ ⇒ ≡ − 2

( )2cos t 1 sin t= − Use of trig identity to deduce that

( )2cos t 1 sin t= − .M1

3 1

2 2gives y sin t cos t= +

Hence 6 6

y sin t cos cos t sinπ π= + = ( )6sin t π+

Using the compound angle

formula to prove y = ( 6sin t )π+ A1 cso

[3]

9 marks


13

Question Number

Scheme Marks

5. (a) Equating i ; 0 6 = + λ 6⇒ λ = − 6λ = − B1 d⇒ Can be implied

Using and 6λ = −

equating j ; a = 19 4( 6) 5+ − = −For inserting their stated into

either a correct j or k component

λ

Can be implied.

M1 d⇒

equating k ; b = 1 2( 6) 11− − − = a 5 and b 11= − = A1

[3]

With no working… … only one of a or b stated correctly gains the first 2 marks.

… both a and b stated correctly gains 3 marks.

(b) ( ) ( ) ( )OP 6 19 4 1 2= + λ + + λ + − − λi juuur

k

direction vector or l1 4 2= = + −d i j k

1OP l OP 0⊥ ⇒ • =duuur uuur

Allow this statement for M1

ifOP and duuur

are defined as above.

ie.

6 1

19 4 4 0

1 2 2

+ λ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟+ λ • =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− − λ −⎝ ⎠ ⎝ ⎠

( )or x 4y 2z 0+ − = Allow either of these two underlined statements

M1

6 4(19 4 ) 2( 1 2 )∴ + λ + + λ − − − λ = 0 Correct equation A1 oe

6 76 16 2 4 0+ λ + + λ + + λ = Attempt to solve the equation in λ dM1

21 84 0 4λ + = ⇒ λ = − 4λ = − A1

( ) ( ) ( )OP 6 4 19 4( 4) 1 2( 4)= − + + − + − − −i j

uuurk

Substitutes their into an

expression for

λOPuuur M1

OP 2 3 7= + +i j k

uuur 2 3 7+ +i j k or P (2, 3, 7) A1

[6]

Note: A similar method may be used by using ( ) ( ) ( )OP 0 5 4 11 2= + λ + − + λ + − λi j kuuur

4 2and = + −j kd i

OP 0• =duuur

yields 6 4( 5 4 ) 2(11 2 ) 0+ λ + − + λ − − λ =This simplifies to . 21 42 0 2λ − = ⇒ λ =

( ) ( ) ( )OP 0 2 5 4(2) 11 2(2)= + + − + + −i juuur

k

OP 2 3 7= + +i j kuuur


14

Question Number

Scheme Marks

Aliter

(b) ( ) ( ) ( )OP 6 19 4 1 2= + λ + + λ + − − λi juuur

k

Way 2

( ) ( ) ( )AP 6 0 19 4 5 1 2 11= + λ − + + λ + + − − λ −i juuur

k

direction vector or l1 4 2= = + −d i j k

AP OP AP OP 0⊥ ⇒ • =uuur uuur uuur uuur

Allow this statement

for M1 if AP and OPuur ru uuu

are defined as above.

ie.

6 6

24 4 19 4 0

12 2 1 2

+ λ + λ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟+ λ • + λ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− − λ − − λ⎝ ⎠ ⎝ ⎠

underlined statement M1

( 6 )(6 ) (24 4 )(19 4 ) ( 12 2 )( 1 2 ) 0∴ + λ + λ + + λ + λ + − − λ − − λ = Correct equation A1 oe

2 236 12 456 96 76 16 12 24 2 4 0+ λ + λ + + λ + λ + λ + + λ + λ + λ =2

Attempt to solve the equation in λ

dM1

221 210 504 0λ + λ + =

( )2 10 24 0 6 4λ + λ + = ⇒ λ = − λ = − 4λ = − A1

( ) ( ) ( )OP 6 4 19 4( 4) 1 2( 4)= − + + − + − − −i juuur

k

Substitutes their λ

into an expression for

OPuuur M1

OP 2 3 7= + +i j k

uuur 2 3 7+ +i j k or

P(2, 3, 7)

A1

[6]

Note: A similar method to way 2 may be used by using ( ) ( ) ( )OP 5 15 4 1 2= + λ + + λ + − λi j kuuur

and ( ) ( ) (AP 5 0 15 4 5 1 2 11= + λ − + + λ + + − λ −i juuur

)kAP OP 0• =uuur uuur

yields ( 5 )(5 ) (20 4 )(15 4 ) ( 10 2 )(1 2 ) 0+ λ + λ + + λ + λ + − − λ − λ =

This simplifies to . 221 168 315 0λ + λ + = ( )2 8 15 0 5λ + λ + = ⇒ λ = − λ = − 3

k

( ) ( ) ( )OP 5 3 15 4( 3) 1 2( 3)= − + + − + − −i juuur

OP 2 3 7= + +i j kuuur


15

Question Number

Scheme Marks

5. (c) OP 2 3 7= + +i j kuuur

OA 0 5 11= − +i j kuuur

and OB 5 15= + +i juuur

k

( )AP 2 8 4=± + −i j kuuur

, ( )PB 3 12 6=± + −i j kuuur

( )AB 5 20 10= ± + −i j kuuur

Subtracting vectors to find any two

of APuuur

, PBuuur

or ABuuur

; and both are correctly ft using candidate’s

OAuuur

andOPuuur

found in parts (a) and

(b) respectively.

M1;

A1 ±

As ( )2 2

3 3AP 3 12 6 PB= + − =i j kuuur uuur

or ( )5 52 2

AB 2 8 4 AP= + − =i j kuuur uuur

or ( )5 53 3

AB 3 12 6 PB= + − =i j kuuur uuur

or ( )3 32 2

PB 2 8 4 AP= + − =i j kuuur uuur

or ( )2 25 5

AP 5 20 10 AB= + − =i j kuuur uuur

or ( )3 35 5

PB 5 20 10 AB= + − =i j kuuur uuur

etc…

alternatively candidates could say for example that

( )AP 2 4 2= + −i j kuuur

( )PB 3 4 2= + −i j kuuur

then the points A, P and B are collinear.

23

AP PB=uuur uuur

or 52

AB AP=uuur uuur

or 53

AB PB=uuur uuur

or 32

PB AP=uuur uuur

or 25

AP AB=uuur uuur

or 35

PB AB=uuur uuur

A, P and B are collinearCompletely correct proof.

A1

AP : PB 2 : 3∴ =uuur uuur

2:3 or 32

1: or 84 : 189 aef B1 oe

allow SC 23

[4]

Aliter

5. (c) At B; 5 6 , 15 19 4 or 1 1 2= + λ = + λ = − − λ

or at B; 1λ = −Writing down any of the three

underlined equations.M1

Way 2

gives for all three equations. 1λ = −or when , this gives 1λ = − =r 5 15+ +i j k

1λ = − for all three equations

or 1λ = − gives =r 5 15+ +i j kA1

Hence B lies on l1. As stated in the question both A and P lie on l1. ∴ A, P and B are collinear.

Must state B lies on l1 ⇒A, P and B are collinear

A1

AP : PB 2 : 3∴ =

uuur uuur 2:3 or aef B1 oe

[4]

13 marks

Beware of candidates who will try to fudge that one vector is multiple of another for the final A mark in part (c).


16

Question Number

Scheme Marks

6. (a)

x 1 1.5 2 2.5 3

y 0 0.5 ln 1.5 ln 2 1.5 ln 2.5 2 ln 3

or y 0 0.2027325541… ln2 1.374436098… 2 ln 3

Either 0.5 ln 1.5 and 1.5 ln 2.5or awrt 0.20 and 1.37

B1

(or mixture of decimals and ln’s) [1]

(b)(i) ( ) }{1

1I 1 0 2 ln2 2ln3

2≈ × × + +

For structure of trapezium

rule{ }............. ; M1;

1

3.583518938... 1.791759...2

= × = = 1.792 (4sf) 1.792 A1 cao

(ii)

( }{ )2

1I 0.5 ; 0 2 0.5ln1.5 ln2 1.5ln2.5 2ln3

2≈ × × + + + +

Outside brackets 1

0.52×

For structure of trapezium

rule{ }............. ;

B1;

M1

1

6.737856242... 1.684464...4

= × = awrt 1.684 A1

[5]

(c) With increasing ordinates, the line segments at the top of the trapezia are closer to the curve.

Reason or an appropriate diagram elaborating the correct reason.

B1

[1]

Beware: In part (b) candidate can add up the individual trapezia:

(b)(i) ( ) ( )1 11 2 2I 0 ln2 ln2 ln3≈ + + +

(ii) ( ) ( ) ( ) ( )1 1 1 1 1 1 1 12 2 2 2 2 2 2 2 2I . 0 0.5ln1.5 . 0.5ln1.5 ln2 . ln2 1.5ln2.5 . 1.5ln2.5 2ln3≈ + + + + + + +


17

Question Number

Scheme Marks

6. (d) 2

du 1dx x

dv xdx 2

u ln x

x 1 v x

= ⇒ =⎧ ⎫⎪ ⎪⎨ ⎬

= − ⇒ = −⎪ ⎪⎩ ⎭

Use of ‘integration by parts’ formula in the

correct directionM1

2 2x 1 x

I x ln x x2 x 2

⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ dx Correct expression A1

2x x

x ln x 1 dx2 2

⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ∫

An attempt to multiply at least one term through by

1x

and an attempt to ...

2 2x x

x ln x x2 4

⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠

⎟ (+c) … integrate;

correct integration

M1; A1

∴3

2 2

1

x xI x ln x

2 4

⎡ ⎤⎛ ⎞= − − +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦x

( ) ( )3 9 1 12 4 2 4ln3 3 ln1 1= − + − − − + Substitutes limits of 3 and

1 and subtracts. ddM1

3 3 342 4

ln3 0= + + − 32ln3= AG 3

2ln3 A1 cso

[6]

Aliter

6. (d) (x 1)ln x dx x ln x dx ln x dx− = −∫ ∫ ∫

Way 2 2 2x x 1

x ln x dx ln x . dx2 2 x

⎛ ⎞= − ⎜ ⎟⎝ ⎠∫ ∫

Correct application of ‘by parts’

M1

=2 2x

ln x2 4

−x

(+ c) Correct integration A1

1

ln x dx x ln x x. dxx

⎛ ⎞= − ⎜ ⎟⎝ ⎠∫ ∫

Correct application of ‘by parts’

M1

= x ln x x− (+ c) Correct integration A1

( ) ( ) ( )3

92

1

x 1 ln x dx ln3 2 3ln3 2∴ − = − − −∫ 32ln3= AG

Substitutes limits of 3 and 1 into both integrands and

subtracts. 32ln3

ddM1 A1 cso

[6]


18

Question Number

Scheme Marks

Aliter

6. (d) ( ) ( )2

du 1dx x

x 1dvdx 2

u ln x

x 1 v−

= ⇒ =⎧ ⎫⎪ ⎪⎨ ⎬

= − ⇒ =⎪ ⎪⎩ ⎭

Use of ‘integration by parts’ formula in the

correct directionM1

Way 3 ( ) ( )2 2

x 1 x 1I ln x

2 2x

− −= − ∫ dx Correct expression A1

( )22x 1 x 2x 1

ln x dx2 2x

− − += − ∫

Candidate multiplies out numerator to obtain three

terms…

( )2

x 1 1 1ln x x 1 dx

2 2 2x

− ⎛ ⎞= − − +⎜ ⎟⎝ ⎠∫

… multiplies at least one

term through by 1x

and

then attempts to ...

( )22x 1 x 1

ln x x ln x2 4 2

− ⎛ ⎞= − − +⎜ ⎟

⎝ ⎠ (+c)

… integrate the result;

correct integration

M1; A1

∴( )

32

2

1

x 1 x 1I ln x x l

2 4 2

⎡ ⎤−⎢ ⎥= − + −⎢ ⎥⎣ ⎦

n x

( ) ( )9 1 14 2 4

2ln3 3 ln3 0 1 0= − + − − − + − Substitutes limits of 3 and 1 and subtracts.

ddM1

31 12 4 4

2ln3 ln3 1= − + + − 32ln3= AG 3

2ln3 A1 cso

[6]

Beware: 1

dx2x∫ can also integrate to

1ln2x

2

Beware: If you are marking using WAY 2 please make sure that you allocate the marks in the order they appear on the mark scheme. For example if a candidate only integrated lnx correctly then they would be awarded M0A0M1A1M0A0 on ePEN.


19

Question Number

Scheme Marks

Aliter By substitution

6. (d) du 1dx x

u ln x= ⇒ =

Way 4 ( )u uI e 1 .ue d= −∫ u Correct expression

( )2u uu e e du= −∫ Use of ‘integration by

parts’ formula in the correct direction

M1

2u u 2u u1 1u e e e e dx

2 2

⎛ ⎞ ⎛= − − −⎜ ⎟ ⎜⎝ ⎠ ⎝∫

⎞⎟⎠

Correct expression A1

2u u 2u u1 1u e e e e

2 4

⎛ ⎞ ⎛= − − −⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠

(+c) Attempt to integrate;

correct integration

M1; A1

∴ln3

2u u 2u u

ln1

1 1I ue ue e e

2 4

⎡ ⎤= − − +⎢ ⎥⎣ ⎦

( ) ( )9 9 12 4ln3 3ln3 3 0 0 1= − − + − − − +

4 Substitutes limits of ln3

and ln1 and subtracts. ddM1

3 3 12 4 4ln3 1= + + − 3

2ln3= AG 3

2ln3 A1 cso

[6]

13 marks


20

Question Number

Scheme Marks

7. (a) From question, dS

8dt

= dS

8dt

= B1

2 dS

S 6x 12xdx

= ⇒ = dS

12xdx

= B1

23dx dS dS 8

;dt dt dx 12x x

= ÷ = = ( )23

k⇒ = Candidate’s dS dS

dt dx÷ ;

8

12x M1; A1oe

[4]

(b) 3 2dV

V x 3xdx

= ⇒ = 2dV3x

dx= B1

2dV dV dx 2

3x . ; 2xdt dx dt 3x

⎛ ⎞= × = =⎜ ⎟⎝ ⎠

Candidate’s dV dx

dx dt× ; xλ

M1;

A1

As 13x V= , then

13

dV2V

dt= AG Use of

13x V= , to give

13

dV2V

dt= A1

[4]

(c) =∫ ∫13

dV2 dt

V

Separates the variables with

13

dV

V∫ or 13V dV

−∫ on one side and

2 dt∫ on the other side.

B1

integral signs not necessary. − =∫ ∫

13V dV 2 dt

Attempts to integrate and … 2

332

V 2t= (+c) … must see 23V and 2t;

Correct equation with/without + c.

M1; A1

233

2(8) 2(0) c c 6= + ⇒ =

Use of V = 8 and t = 0 in a changed equation containing c ;

c 6=M1∗ ; A1

Hence:

233

2V 2t= + 6

Having found their “c” candidate … ( )

233

216 2 2t 6= + 12 ⇒ 2t 6= + … substitutes V 16 2= into an

equation involving V, t and “c”. depM1∗

giving t = 3. t = 3 A1 cao [7] 15 marks


21

Question Number

Scheme Marks

Aliter

7. (b) 1 23 32x V & S 6x S 6V= = ⇒ =

23S 6V= B1

Way 2

13

dS4V

dV

−= or 13

dV 1V

dS 4=

13

dS4V

dV

−= or 13

dV 1V

dS 4= B1

13

1 13 3

dV dS dV 1 28. ; 2V

dt dt dS 4V V− −

⎛ ⎞= × = = =⎜ ⎟⎝ ⎠

AG Candidate’s dS dV

dt dS× ;

132V M1; A1

In ePEN, award Marks for Way 2 in the order they appear

on this mark scheme. [4]

Aliter

7. (c) 13

dV1 dt

2V=∫ ∫


13

dV

2V∫ or 13

1V dV

2

−∫ oe on one

side and 1 dt∫ on the other side.

B1

Way 2 integral signs not necessary. 1

31

V dV 1 dt2

− =∫ ∫

Attempts to integrate and … ( ) ( ) 2

3312 2

V t= (+c) … must see 23V and t;


M1; A1

233

4(8) (0) c c 3= + ⇒ =


c 3=M1∗ ; A1

Hence:

233

4V t= + 3

Having found their “c” candidate …

( )233

416 2 t 3= + 6 t ⇒ 3= + … substitutes V 16 2= into an


giving t = 3. t = 3 A1 cao [7]

Beware: On ePEN award the marks in part (c) in the order they appear on the mark scheme.


22

Question Number

Scheme Marks

Aliter similar to way 1.

(b) 3 2dV

V x 3xdx

= ⇒ = 2dV3x

dx= B1

Way 3

2dV dV dS dx 1

3x .8. ; 2xdt dx dt dS 12x

⎛ ⎞= × × = =⎜ ⎟⎝ ⎠

Candidate’s dV dS dx

dx dt dS× × ; xλ

M1;

A1

As 13x V= , then

13

dV2V

dt= AG Use of

13x V= , to give

13

dV2V

dt= A1

[4]

Aliter

(c) =∫ ∫13

dV2 dt

V


13

dV

V∫ or 13V dV

−∫ on one side and

2 dt∫ on the other side.

B1

Way 3 integral signs not necessary. − =∫ ∫

13V dV 2 dt

Attempts to integrate and …

23 4

3V = t (+c)

… must see 23V and 4

3t;


M1; A1

23 4

3(8) (0) c c 4= + ⇒ =


c 4=M1∗ ; A1

Hence:

23 4

3V t= + 4

Having found their “c” candidate … ( )

23 4

316 2 t 6= + ⇒ 4

38 t= + 4 … substitutes V 16 2= into an


giving t = 3. t = 3 A1 cao [7]

• Beware when marking question 7(c). There are a variety of valid ways that a candidate can use to find the constant “c”.

• In questions 7(b) and 7(c) there may be “Ways” that I have not listed. Please use the mark scheme as a guide of how the mark the students’ responses.

• In 7(c), if a candidate instead tries to solve the differential equation in part (a) escalate the response to your team leader.

• IF YOU ARE UNSURE ON HOW TO APPLY THE MARK SCHEME PLEASE ESCALATE THE RESPONSE UP TO YOUR TEAM LEADER VIA THE REVIEW SYSTEM.

• Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark. ddM1 denotes a method mark which is dependent upon the award of the previous two method marks.

depM1∗ denotes a method mark which is dependent upon the award of M1 . ∗

core mathematics c4 (6666) level... · 6666/01 core maths c4 june 2006 advanced subsidiary/...

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